Looking at various recent examination papers, it has become clear to
me that there is significant confusion between these two questions.
This post is intended to bring some clarity to the situation.

At the start of this post, I will give an example of the confusion as
it appears in exam questions (and probably elsewhere), and clarify
what the two different phrases mean using the above example. I will
then delve more deeply into the mathematics of these two things, going
beyond A-level content, and use some undergraduate analysis to find
equivalent conditions for them in terms of the derivatives of the
functions. It is fine to skip over the technical stuff and just look
at the results (theorems)!

(Exactly the same applies to the use of the term “decreasing”, but for
simplicity we will focus on increasing functions in this post.)

Here is an example of a question (based on a real exam question) which
typifies the confusion.

The equation of a curve is $y=x^3+4x^2-5x$.

Find the set of values of $x$ for which $y$ is an increasing
function of $x$.

If we replace “increasing function” by another familiar A-level term
describing functions, “one-to-one function”, the question becomes:

A function is given by $f(x)=x^3+4x^2-5x$.

Find the set of values of $x$ for which $f(x)$ is a one-to-one
function of $x$.

This is clearly nonsensical, because whether a function is one-to-one
or not is a property of the function as a whole, not a property of
the function values at any particular input value.

Likewise, a function either is or is not an increasing function; it
is a property of the function as a whole.

Informally (and not quite correctly), we can describe the difference
as follows:

A function is increasing at a point if at that point, the function
has a positive gradient.

An example which shows that these are not the same is the function
$f(x)=-\dfrac{1}{x}$ for $x\ne0$ shown above. This function is
increasing at every value of $x\ne0$, as the gradient is always
positive. However, it is not an increasing function, because
$f(1)<f(-1)$. If, though, we restricted the domain of the function to
$x>0$, then it would be an increasing function.

So the above-quoted exam question does not make any sense, just as the
modified version did not: either $y$ is an increasing function of $x$
or it is not. If the question had instead asked “Find the set of
values of $x$ at which $y$ is increasing,” it would have been fine.

Incidentally, the idea of increasing and decreasing functions connects
very well with the issue of rearranging inequalities (increasing the
depth of connections within the subject): a function can be applied to
both sides of an inequality without changing the direction of the
inequality if the function is (strictly) increasing; it can be applied
but with a change in the direction of the inequality if the function
is (strictly) decreasing, and if the function is neither, then the
function cannot be applied to the inequality. So we cannot square
both sides of an inequality unless we are restricted to non-negative
values, and we cannot take the reciprocal of an inequality unless we
have the same restriction (and in that case, we must also reverse the
direction of the inequality).

It seems reasonable to assert that if a function is an increasing
function, then it will be increasing at every point. There turns out
to be some subtlety to this, as we now delve into a little more
deeply.

A formal definition of increasing

We can give a formal definition of an increasing function. For
example, this definition is from Apostol, Mathematical Analysis, 2nd
ed, p94, and identical definitions appear on the internet:

Definition 1: Let $f$ be a real-valued function whose domain is a
subset $S$ of $\mathbb{R}$.
Then $f$ is said to be an increasing (or nondecreasing)
function if for every pair of points $x$ and $y$ in $S$,
$x<y$ implies $f(x)\le f(y)$.
If $x<y$ implies $f(x)<f(y)$, then $f$ is said to be a strictly
increasing function. (Decreasing functions are similarly defined.)

Note the distinction between increasing and strictly increasing here:
a constant function such as $f(x)=0$ for $x\in\mathbb{R}$ is both an
increasing and decreasing function, though it is not a strictly
increasing function.

We could also try to come up with a definition of increasing at a
point. There are no standard definitions of this idea, and the
following proposed definition is certainly beyond A-level in its
formality. It is based on the definition of continuity, which is
about the behaviour of a function “near” to a point.

Definition 2: Let $f$ be a real-valued function whose domain is a
subset $S$ of $\mathbb{R}$. Then $f$ is said to be increasing at
the point $x$ in $S$ if there is some $\delta>0$ such that:

for every $y$ in $S$ with $x<y<x+\delta$, $f(x)\le f(y)$, and
for every $y$ in $S$ with $x-\delta<y<x$, $f(y)\le f(x)$.

If the $\le$ signs are replaced by $<$ signs in these two
inequalities, then $f$ is said to be strictly increasing at $x$.

With this definition, the above exam question (reworded) makes sense,
and the correct final answer is what the examiner would expect. (One
might wonder whether one could make such a local definition of
one-to-one, and indeed, this is done when considering the Inverse
Function and Implicit Function theorems. But that is a story for
another day.)

Using calculus

So far, no calculus has appeared, yet we typically teach our students
to determine whether a function is an increasing function or to find
where it is increasing by differentiating the function. So let us now
consider how we could use calculus to help us.

For us to be able to use calculus, we need to assume that our function
is differentiable throughout $S$. We could then propose the following
theorem:

Theorem 1 (incorrect attempt): Let $f$ be a real-valued continuous
function whose domain is a subset $S$ of $\mathbb{R}$ and is
differentiable at every (interior) point of $S$. Then $f$ is an
increasing function if and only if $f’(x)>0$ for all $x$ in (the
interior of) $S$.

We could try changing this to say that $f$ is a strictly increasing
function, but that fails if the function has a point of inflection.
For example, $f(x)=x^3$ is a strictly increasing function, even though
its derivative is zero at $x=0$.

We could also try changing the condition to say that $f’(x)\ge0$ for
all $x$ in $S$. However, this also fails: if the graph has a
discontinuity, such as the function $f(x)=-\dfrac{1}{x}$ for $x\ne0$
that we looked at before, then it might have $f’(x)>0$ for all $x$ in
$S$, yet not be an increasing function.

This feels more hopeful, though: after all, the only problem now is
the “hole” in the domain $S$. And it turns out that if we restrict
the domain to be an interval (that is, a subset of the reals with no
“holes”), then it will work:

Theorem 1 (correct version): Let $f$ be a real-valued continuous
function whose domain is an interval $I$ of $\mathbb{R}$ and is
differentiable at every point in (the interior of) $I$. Then $f$ is
an increasing function if and only if $f’(x)\ge 0$ for all $x$ in
(the interior of) $I$.

The formal proof of this is found below, and though it is quite
technical, the theorem itself seems clearly true, and school students
could probably be convinced to believe it (at least once it is written
in more student-friendly language).

What can we say, though, about whether a (differentiable) function is
increasing at a point? Using Definition 2 above, we get the
corresponding theorem:

Theorem 2: Let $f$ be a real-valued continuous function whose domain
is an interval $I$ of $\mathbb{R}$ and which is differentiable at
every (interior) point of $I$. Then is $f$ is increasing at the
point $x$ in $I$ if and only if there is some $\delta>0$ for which
$f’(y)\ge0$ for all $y$ in (the interior of) $I$ with
$x-\delta<y<x+\delta$.

Why is it not sufficient to just require $f’(x)\ge0$? Well, consider
the functions $f(x)=x^3$ and $f(x)=-x^3$. They both have $f’(x)=0$,
yet the first is increasing (indeed, even strictly increasing) at
$x=0$, while the second is decreasing at $x=0$. And a function such
as $f(x)=x^2$ is neither increasing nor decreasing at $x=0$. So we
really do need to consider a small interval around the point of
interest.

(Theorem 2 could be extended, with care, to more general subsets of
$\mathbb{R}$, as we are only discussing a local property of the
function. But it is not particularly interesting to do so.)

So the question of determining at which points a function is
increasing (or decreasing) is more subtle than it appears: not only
does one have to find where the function has derivative $\le0$ (and
not just $<0$), but one also has to determine what is happening at
those points where the derivative is zero, as there are different
types of stationary points. (At those points where the derivative is
strictly positive, the function is certainly strictly increasing,
which follows from Theorem 4 below.)

Things get more complicated if we now wish to consider strictly
increasing (or decreasing) functions. There is a relatively weak
theorem which will suffice much of the time:

Theorem 3: Let $f$ be a continuous real-valued function whose domain
is an interval $I$ of $\mathbb{R}$ and which is differentiable at
every (interior) point of $I$. Then if $f’(x)>0$ throughout $I$,
$f$ is a strictly increasing function.

Note that this is a one-directional theorem; $f(x)=x^3$ for
$x\in\mathbb{R}$ is our standard example of a strictly increasing
function which does not have $f’(x)>0$ throughout the domain because
of the point of inflection at the origin. The proof of Theorem 3
follows exactly as that of Theorem 1.

An easy corollary of this is the following (local) theorem:

Theorem 4: Let $f$ be a continuous real-valued function whose domain
is a subset $S$ of $\mathbb{R}$. If $f$ is differentiable at the
point $x$ in the interior of $S$ and $f’(x)>0$, then $f$ is strictly
increasing at $x$.

This is the theorem which is typically used when answering A-level
exam questions such as the one above. Unfortunately, as we see from
our example of $f(x)=x^3$, this too is a one-directional theorem:
every point at which $f’(x)>0$ is a point at which the function is
strictly increasing, but there may be other points where this is the
case but where $f’(x)=0$. (If $f’(x)<0$, then the function is
strictly decreasing at this point, so it cannot be increasing.) The
question of using calculus to determine where a function is
increasing, rather than strictly increasing, is somewhat more
complicated, as we see from Theorem 2 above. But at A-level, the
functions are always nice enough that the only difficulties will be at
the stationary points.

There is actually a necessary and sufficient condition for a function
to be strictly increasing, but this is more subtle:

Theorem 5: Let $f$ be a continuous real-valued function whose domain
is an interval $I$ of $\mathbb{R}$ and which is differentiable at
every interior point of $I$. Then $f$ is strictly increasing on $I$
if and only if $f’(x)\ge0$ throughout $I$ and there is no
non-trivial subinterval $J$ of $I$ with $f’(x)=0$ for all $x$ in the
interior of $J$.

The proof can be found below.

Teaching this topic at A-level

Putting this all together, we see that Theorem 4 is the crucial
theorem for school use. Teaching the meaning of the term “increasing
function” (Definition 1) and a simplified explanation of “increasing
at a point” (Definition 2), along with Theorem 4 should give a good
grounding. It would also be wise to caution that it is a one-way
theorem by comparing and contrasting examples such as $f(x)=x^2$ and
$f(x)=x^3$.

Proofs of Theorems 1 and 5

This technical appendix uses tools from undergraduate analysis. The
proofs of the other three theorems are very similar to these or they
follow immediately from these.

Theorem 1

Let $f$ be a real-valued continuous function whose domain is an
interval $I$ of $\mathbb{R}$ and is differentiable at every point in
the interior of $I$. Then $f$ is an increasing function if and only
if $f’(x)\ge 0$ for all $x$ in the interior of $I$.

Proof

We show first that if $f$ is an increasing function, then $f’(x)\ge0$
for all $x$ in the interior of $I$, and we argue by contradiction.
Assume that $f’(x_0)<0$ for some $x_0$ in the interior of $I$. Using
the definition of derivative, this means that
$\lim\limits_{\substack{x\to x_0\\ x\in
I}}\dfrac{f(x)-f(x_0)}{x-x_0}<0$. So there is some $x_1\in I$ (where
$x_1\ne x_0$) with $\dfrac{f(x)-f(x_0)}{x-x_0}<0$ (otherwise the limit
would be $\ge0$). If $x_1>x_0$, then multiplying by $x_1-x_0$ gives
$f(x_1)-f(x_0)<0$, so $f(x_1)<f(x_0)$, If $x_1<x_0$, then multiplying
by $x_1-x_0$ gives $f(x_1)-f(x_0)>0$, so $f(x_1)>f(x_0)$. Either way,
this shows that the function is not increasing on $I$, and we have our
desired contradition. Thus if $f$ is an increasing function, we must
have $f’(x)\ge0$ for all $x$ in the interior of $I$.

Conversely, if $f’(x)\ge0$ for all $x$ in the interior of $I$, then
let $x<y$ be any two points in $I$. Then by the mean-value theorem,
there is some $z$ with $x<z<y$ for which $f(y)-f(x)=f’(z)(y-x)$ (and
note that $z$ lies in the interior of $I$ as $I$ is an interval).
Since $f’(z)\ge0$ by assumption, and $y-x>0$, it follows that
$f(y)-f(x)\ge0$, so $f(x)\le f(y)$. Therefore $f$ is an increasing
function.

Theorem 5

Let $f$ be a continuous real-valued function whose domain is an
interval $I$ of $\mathbb{R}$ and which is differentiable at every
interior point of $I$. Then $f$ is strictly increasing on $I$ if
and only if $f’(x)\ge0$ throughout $I$ and there is no non-trivial
subinterval $J$ of $I$ with $f’(x)=0$ for all $x$ in the interior of
$J$.

Proof

We first prove that if the derivative condition is not met, then $f$
is not strictly increasing on $I$. If $f’(x)<0$ at any point in $I$,
then $f$ is not increasing (by Theorem 1), so it is certainly not
strictly increasing. If $f’(x)\ge0$ throughout $I$ but there is a
non-trivial subinterval $J$ of $I$ with $f’(x)=0$ for all $x$ in the
interior of $J$, then $f$ is constant throughout $J$ (by the
mean-value theorem). In particular, there are $y<z$ in $J$ with
$f(y)=f(z)$, showing that $f$ is not strictly increasing.

Conversely, if $f’(x)\ge0$ throughout $I$, then $f$ is increasing by
Theorem 1. Assume now that there is no non-trivial subinterval $J$ of
$I$ with $f’(x)=0$ for all $x$ in the interior of $J$. But if $f$
were not strictly increasing, then there would be $y<z$ in $I$ with
$f(y)=f(z)$, so $f(x)$ is constant on the interval $y<x<z$. (For if
$f(y)<f(x)$ for some $x$ in this interval, we would have $f(x)>f(z)$,
contradicting $f$ increasing.) Therefore $f’(x)=0$ throughout this
interval, contradicting our assumption. So $f$ must be strictly
increasing.